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If $d \mid a$ and $d \mid b$ then $d \mid \gcd(a, b)$. Furthermore, $d \mid a-b$ and $d \mid a+b$ so $d \mid \gcd(a-b,a+b)$. This doesn't seem to give any relation directly between the two gcds.

However, in the proof that $a \perp b \implies \gcd(a+b, a-b) = 1$ or $2$, it seems that the conclusion $d \mid \gcd(2a, 2b) = 2$ (where $d$ is defined as a common divisor of $a+b$ and $a-b$) is taken to imply that $\gcd(a+b, a-b)$ also divides 2. This would be true if $d \mid a+b$ and $d \mid a-b$ were to imply that $\gcd(a+b, a-b) \mid d$, but this is the exact opposite of the actual fact that $d \mid \gcd(a+b, a-b)$. So how is this step explained in the $a+b$ and $a-b$ proof and does this generalise to imply the titular result?

iTechnical
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    By the gcd universal property in the first linked dupe $,d=(a,b)\mid a-b,a+b\Rightarrow d\mid(a-b,a+b).,$ Alternatively see the more precise result in the 2nd dupe. – Bill Dubuque Dec 30 '22 at 15:33

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It is true, let $d = \gcd(a,b)$ then $d|a$ and $d|b$ so $d|(a+b)$ and $d|(a-b)$. Thus $d|\gcd(a+b,a-b)$.

Whith regards to the rest of your question, if $a,b$ are coprime then by the above $\gcd(a+b,a-b)|\gcd(2a,2b) = 2$ (replace $a,b$ with $a+b,\ a-b$).

GoodThanks
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